document.write( "Question 1142720: A spherical ballon is being filled with air at a constant rate of 2cm^3 per second. By using the chain rule dr/dt equals to 1/2(pi)(r^2) where t is in seconds. The rate at which the radius is increasing when r=54 is 1/5832pi. The balloon will burst when the radius reaches 100cm. Find the rate at which the surface area is increasing at that point in time. \n" ); document.write( "
Algebra.Com's Answer #763449 by Alan3354(69443)\"\" \"About 
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A spherical ballon is being filled with air at a constant rate of 2cm^3 per second. By using the chain rule dr/dt equals to 1/2(pi)(r^2) where t is in seconds. The rate at which the radius is increasing when r=54 is 1/5832pi. The balloon will burst when the radius reaches 100cm. Find the rate at which the surface area is increasing at that point in time.
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\n" ); document.write( "\"V+=+4%2Api%2Ar%5E3%2F3\"
\n" ); document.write( "dV/dt = 4pi*r^2*dr/dt = 2
\n" ); document.write( "dr/dt = 1/(2pi*r^2) which is 1/(5832pi) at r = 54
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\n" ); document.write( "dr/dt at r=100 is 1/(2pi*10000) = 1/(20000pi)
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\n" ); document.write( "SA = 4pi*r^2
\n" ); document.write( "dSA/dt = 8pi*r*(dr/dt)
\n" ); document.write( "dSA/dt = 8pi*100/20000pi
\n" ); document.write( "= 1/25 sq cm/sec
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\n" ); document.write( "\"...at that point in time.\" is superfluous.
\n" ); document.write( "\"at that point.\" or \"at that time.\" is better and sufficient.
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\n" ); document.write( "No need for \"at the point in space, or in the space-time continuum\" either.\r
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